What is the solution for y^{''}-3y^'+2y=14sin(2x)-18cos(2x) ?

The solution for y^{''}-3y^'+2y=14sin(2x)-18cos(2x) is y=c_{1}e^{2x}+c_{2}e^x+2sin(2x)+3cos(2x)

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y^{''}-3y^'+2y=14sin(2x)-18cos(2x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 14 sin (2 x) - 18 cos (2 x)

y = c_{1} e^{2 x} + c_{2} e^{x} + 2 sin (2 x) + 3 cos (2 x)

Solution steps

y^{''} (x) - 3 y^{'} (x) + 2 y = 14 sin (2 x) - 18 cos (2 x)

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + 2 sin (2 x) + 3 cos (2 x)

y = c_{1} e^{2 x} + c_{2} e^{x} + 2 sin (2 x) + 3 cos (2 x)

\frac{d}{d x} (x + sin (y))

\int \frac{9 x ^{3} - 6 x + 4}{x ^{3} - x ^{2}} d x

\frac{\partial}{\partial t} (e^{- 2 t} cos (2 x))

\int (2 e^{3 x}) d x

\int (4 x^{2} + \frac{1}{( 4 x ) ^{2}}) d x

What is the solution for y^{''}-3y^'+2y=14sin(2x)-18cos(2x) ?
The solution for y^{''}-3y^'+2y=14sin(2x)-18cos(2x) is y=c_{1}e^{2x}+c_{2}e^x+2sin(2x)+3cos(2x)